# Math - Equations of planes

Write a vector equation and scalar equation of the plane containing lines
[x,y,z]= [-2,3,12]+s[-2,1,5]
[x,y,z]= [1,-4,4]+t[-6,3,15]

1. 0
2. 0
3. 13
1. Did you notice that our two lines have the same direction?
(the second vector is a multiple of the first)

So we need a second direction vector, we can obtain that by using the two given points (-2,3,12) and (1,-4,4)
that direction vector is [3,-7,-8]

so a vector equation is
[x,y,z] = (-2,3,12) + s[-2,1,5] + t[3,-7,-8]

for the scalar equation we will take the cross-product of our two direction vectors to get a normal, which is
[27,-1,11]
http://www.analyzemath.com/vector_calculators/vector_cross_product.html

so our equation is
27x - y + 11z = c
plug in (1,-4,4) to get c
27 + 4 +44 = c = 75

So the scalar equation is
27x – y + 11z = 75

1. 0
2. 0
posted by Reiny

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